Ex 1.3, 8 - f(x) = x2 + 4. Show that f is invertible - Chapter 1

Ex 1.3 , 8 - Chapter 1 Class 12 Relation and Functions - Part 2
Ex 1.3 , 8 - Chapter 1 Class 12 Relation and Functions - Part 3
Ex 1.3 , 8 - Chapter 1 Class 12 Relation and Functions - Part 4
Ex 1.3 , 8 - Chapter 1 Class 12 Relation and Functions - Part 5 Ex 1.3 , 8 - Chapter 1 Class 12 Relation and Functions - Part 6

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Ex 1.3, 8 Consider f: R+ → [4, ∞) given by f(x) = x2 + 4. Show that f is invertible with the inverse f−1 of given f by f-1 (y) = √(y−4) , where R+ is the set of all non-negative real numbers. f(x) = x2 + 4 f is invertible if f is one-one and onto Checking one-one f (x1) = (x1)2 + 4 f (x2) = (x2)2 + 4 Putting f (x1) = f (x2) (x1)2 + 4 = (x2)2 + 4 Rough One-one Steps: 1. Calculate f(x1) 2. Calculate f(x2) 3. Putting f(x1) = f(x2) we have to prove x1 = x2 (x1)2 = (x2)2 ⇒ x1 = x2 or x1 = –x2 Since f: R+ → [4,∞ ) So x ∈ R+ i.e. x is always positive, Hence x1 = –x2 is not true So, x1 = x2 ∴ f is one-one Check onto f(x) = x2 + 4 Let f(x) = y such that y ∈ [4,∞ ) y = x2 + 4 y – 4 = x2 x2 = y – 4 x = ±√(𝑦−4) Since f: R+ → [4,∞ ) x ∈ R+ , so x is positive Hence we cannot take x = −√(𝑦−4) x = √(𝑦−4) Since, y ∈ [4,∞ ) i.e. y is greater than or equal to 4 i.e. y ≥ 4 y – 4 ≥ 0 Hence the value inside root is positive Hence, √(𝑦−4) ≥ 0 x ≥ 0 Hence x is a real number which is greater than or equal to 0. ∴ x ∈ R+ Now, Checking for y = f(x) Putting value of x in f(x) f(x) = f(√(𝑦−4)) = (√(𝑦−4))^2+4 = y − 4 + 4 = y Thus, for every y ∈ [4,∞ ) , there exists x ∈ R+ such that f(x) = y Hence, f is onto Since f(x) is one-one and onto, So, f(x) is invertible And Inverse of x = 𝑓^(−1) (𝑦) = √(𝒚−𝟒)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo